Lecture 13: Calcination Contents Principles of calcination Applications Materials and heat balance in calcination Problem – i Decomposition of CaCO Problem – ii amount of fuel in calcination Problem – iii calcination of Al OH Problem – iv fuel saving Conclusions References Key words:calcination, electrolysis of alumina, fuel saving Principles of calcination: Calcination is a thermal treatment process and applied to ores and other solid materials to bring a) thermal decomposition b) phase transition and c) to remove volatile fractions such as CO , H O Material is heated below the melting point in rotary kiln or fluidized bed reactor. Calcination is done in the solid state. Application: 9 To produce cement from CaCO 9 To cause decomposition of hydrated minerals as in calcination of bauxite to produce refractory grade Al O . 9 To cause decomposition of volatile matter contained in petroleum Coke. 9 To heat treat to effect phase transformation as in devitrification of glass materials. 9 To produce anhydrousAl O for electrolysis of Al O to Al in Hall‐Heroult cell Materials and heat balance in calcination Calcination requires thermal energy. This is illustrated by solving few problems. For calculation, we need several thermo chemical values like heat of formation, specific heat, and heat content. The following thermo chemical values arc used to solve the problems in this lecture: Thermo chemical values: CaCO3=CaO+CO2 ∆H
42750 Kcal/Kg.mol MgCO3=MgO+CO2 ∆H
24250 Kcal/Kg.mol CO+1/2 O2=CO2 ∆H
67900 Kcal/Kg.mol 2Al(OH)3=Al2O3+3H2O ∆H
24290 Kcal/Kg.mol C+O2=CO2 ∆H
94300 Kcal/Kg.mol H2+1/2 O2=H2O ∆H
68370 Kcal/Kg.mol CpCaO=49.622+4.519x10‐3T‐6.945x105 CpMgO=48.995+3.138x10‐3T‐11.715x105
kJ/Kg.mol K kJ/Kg.mol K CpCO2=75.438 kJ/Kg.mol K CpH2O(v)=30.+10.711x10‐3T‐0.335x105
kJ/Kg.mol K Heat content H1200 ‐ H298|CaO =10800 Kcal/Kg.mol H500 ‐ H298|CO2 H500 ‐ H298|N2 H500 ‐ H298|O2 H1000 ‐ H298|Al2O3 =1987 =1418 =1455 =18710 Kcal/Kg.mol Kcal/Kg.mol Kcal/Kg.mol Kcal/Kg.mol H800 ‐ H298|CO2 =5458 Kcal/Kg.mol H800 ‐ H298|O2 H800 ‐ H298|N2 H800 ‐ H298|H2O(l) H900 ‐ H298|CO2 H900 ‐ H298|O2 H900 ‐ H298|N2 =3786 = 3598 = 14824 = 6708 = 4602 = 4358 Kcal/Kg.mol Kcal/Kg.mol Kcal/Kg.mol Kcal/Kg.mol Kcal/Kg.mol Kcal/Kg.mol H900 ‐ H298|CO = 4400 Kcal/Kg.mol Latent Heat of vaporization of water = 10520 kJ/kg.mol Problem‐i: decomposition of 1) Calculate the heat energyrequired to calcine 1000Kg limestone of composition 84%CaCO , 8%MgCO and 8% H O charged at 298K. Lime is discharged at 1173K and gases leave at 473K. Solution: CaCO
Mg CO
CaO
MgO
CO (1) CO (2) Material balance gives Calcined product and gases CaO
8.4 kg mols MgO
0.952 kg mols CO
9.352kg mols H O
4.444 kg mols 382186 k cal. Heat of decomposition of reaction 1 and2 Using Cp values one can calculate sensible heat in calcined products and gases. Reference state is 298K Sensible heat in products
Sensible heat in CO
8.4
9.352
Cp CaO dT
0.952
Cp MgO dT Cp CO dT Sensible heat in H O l can be evaluated as H O
,
H O
,
H O
,
H O
,
H O
.
4.44
Cp H O
dT Answer Problem 2: amount of fuel in calcination It is desired to produce 10 kg. mol lime from calcinations of CaCO (pure) in a rotary kiln. Producer gas of composition CO 7.2%, O 1.6%, CO 16.6% and N 74.6% is combusted with 20% excess air to obtain the desired temperature in the kiln. The limestone and air are supplied at 298K, whereas producer gas is heated to 900K. Lime is discharged at 1200 K and at 500K. Calculate the amount of producer gas (1 atm. and 273 K). Let Y kg mol is the producer gas Material balance gives 10
CO
N
0.238 Y 0.74 Y
O
0.302 Y 0.0134 Y Calorific value of producer gas= 11271 Y kcal. Performing heat balance: Heat input= Heat output. Sensible heat in producer gas+ calorific value of producer gas – Heat of decomposition of CaCO = sensible heat in CaO sensible heat in flue gases CO , N and O We calculate all the values and get .
Answer Problem 3 Calcination of In the electrolysis, anhydrous alumina is required. For this purpose Al OH is calcined at 1700 K in rotary kiln. A kiln receives a damp filter cake of Al OH analyzing 55% Al O and 45% total H O (free and combined) and produce, pure Al O as solid product. The fuel consumption is estimated to be 0.2Kg of fuel oil of composition 84% C and 16% H per Kg of alumina. Air for combustion is 20% excess than theoretical required. Assume complete combustion and heat losses 10% of heat input. Find a) The volume of gases (At 1 atm, 273 K)leaving the kiln per 1000Kg of Al O produced. b) Wet and dry composition of flue gases. c) Perform the heat balance and comment on the results. Assume reactants enter at 298K and products namely Al O at 1000K and flue gases at 800K. Solution: Basis of calculation 1000Kg calcineAl O Calcination reaction: 2Al OH
Al O
3H O Combustion reactions C
2H
Material balance gives volume of flue gases CO O
0.5O
H O On dry basis (%) 11.9 3.7 84.4 ‐ 100% Flue gas analysis: on Wet basis (%) CO2 7.8 O2 2.5 N2 55.4 H2O 34.3 100% Heat balance gives the following result: Heat input (kcal) Combustion of fuel +2414120* Heat of decomposition – 2381370 Heat available: 2175983 Heat output (kcal) Flue gases 1362135 Sensible heat in Al2O3183431 Heat losses 217598 Total 1763164 kcal *
+ indicated heat input due to exothermic and ‐ indicates heat absorption. Heat balance indicates that there is 412819 kcal heat is surplus. This surplus heat may be utilized. If not then amount of fuel may be reduced as illustrated in problem 4 Problem 4: Fuel saving Calculate the minimum amount of fuel 1000Kg Al O . Use the data given in problem3. Let x kg fuel is required. C
0.84 xandH
0.16 x We have to calculate flue gas. The calculation gives kg mols CO
H O
0.07 x 0.08x
45.45 O
0.022 x N
0.496 x Heat balance: Heat of combustion
Heat taken by flue gas
Heat taken by Al O
heat losses
Performing heat balance calculation, we can get x
141kgfuel is required to produce 1000Kg alumina. We save /
Conclusion: This lecture discusses the basics of calcination by solving problems. The importance of heat balance calculations is shown in problem 4 which shows that fuel saving can be achieved. References: 1) Rao, Y.K: Stoichiometry and Thermodynamics of Metallurgical process, Cambridge University Press, 1985 2) Butts, Allison: Metallurgical Problems, McGraw Hill Book Company, 1943 3) Fine, H.Alan and G.H.Geiger: Handbook on Materials and Energy Balance Calculations in Metallurgical Processes